Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
using pii = pair<int, int>;
using vpii = vector<pii>;
const int inf = 1e9L + 7;
const ll INF = 1e18L + 7;
const ll N = 2e5 + 7;
ll sy, a, b, n, q;
ll l[N];
ll r[N];
ll sum[N];
bool prawo(ll x1, ll y1, ll x2, ll y2) { return x1 * y2 - x2 * y1 > 0; }
double query(ll x, ll y) {
  ll lo = 1, hi = n + 1, L = -1, R = -1;
  while (lo < hi) {
    ll m = (lo + hi) / 2;
    if (hi == m) --m;
    if (prawo(a - x, sy - y, r[m] - x, 0 - y))
      hi = m;
    else
      lo = m + 1;
  }
  L = lo;
  lo = 0, hi = n;
  while (lo < hi) {
    ll m = (lo + hi) / 2;
    if (lo == m) ++m;
    if (prawo(l[m] - x, 0 - y, b - x, sy - y))
      lo = m;
    else
      hi = m - 1;
  }
  R = lo;
  if (L > R) return 0;
  if (L == R) {
    double X1 = max((double)l[L],
                    a * (double)y / (y - sy) + x * (double)-sy / (y - sy));
    double X2 = min((double)r[R],
                    b * (double)y / (y - sy) + x * (double)-sy / (y - sy));
    double ans = X2 - X1;
    ans *= (double)(y - sy) / y;
    return ans;
  }
  double ans = sum[R - 1] - sum[L];
  double X1 =
      max((double)l[L], a * (double)y / (y - sy) + x * (double)-sy / (y - sy));
  double X2 =
      min((double)r[R], b * (double)y / (y - sy) + x * (double)-sy / (y - sy));
  ans += r[L] - X1;
  ans += X2 - l[R];
  ans *= (double)(y - sy) / y;
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout << fixed << setprecision(15);
  cin >> sy >> a >> b >> n;
  for (ll i = 1; i <= n; ++i) {
    cin >> l[i] >> r[i];
    sum[i] = sum[i - 1] + r[i] - l[i];
  }
  cin >> q;
  while (q--) {
    ll x, y;
    cin >> x >> y;
    cout << query(x, y) << '\n';
  }
}